Question

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to gravity at the surface of the moon in 1.62 m/s2. What is the maximum height of the rock above the surface?

Answer 1

Answer: 277.777 m

Step-by-step explanation:

The situation described here is parabolic movement. However, as we are told that the rock was projected upward from the surface, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

`y-y_(o)=V_(o).t+(1)/(2)g.t^(2)` (1)

`V=V_(o)-g.t` (2)

Where:

`y` is the rock's final position

`y_(o)=0` is the rock's initial position

`V_(o)=30(m)/(s)` is the rock's initial velocity

`V` is the final velocity

`t` is the time the parabolic movement lasts

`g=1.62(m)/(s^(2))` is the acceleration due to gravity at the surface of the moon

As we know
`y_(o)=0` , equation (2) is rewritten as:

`y=V_(o).t+(1)/(2)g.t^(2)` (3)

On the other hand, the maximum height is accomplished when
`V=0`:

`V=V_(o)-g.t=0` (4)

`V_(o)-g.t=0`

`V_(o)=g.t` (5)

Finding
`t`:

`t=(V_(o))/(g)` (6)

Substituting (6) in (3):

`y=V_(o)((V_(o))/(g))+(1)/(2)g((V_(o))/(g))^(2)` (7)

`y_(max)=\frac{{V_(o)}^(2)}{2g}` (8) Now we can calculate the maximum height of the rock

`y_(max)=\frac{{(30m/s)}^(2)}{(2)(1.62m/s^(2))}` (9)

Finally:

`y_(max)=277.777m`